Second Law of Thermodynamics: Heat Engines and Carnot Cycle, Study notes of Thermodynamics

Download Second Law of Thermodynamics: Heat Engines and Carnot Cycle and more Study notes Thermodynamics in PDF only on Docsity! Atmospheric Thermodynamics The Second Law of Thermodynamics and Entropy The first law of thermodynamics is a statement of conservation of energy. The second law of thermodynamics is concerned with the maximum fraction of a quantity of heat that can be converted into work. The Carnot Cycle Cyclic process - a series of operations by which the state of a substance (called the working substance) changes but the substance is finally returned to its original state in all respects Work (w) is done by (or on) the working substance if its volume changes. The internal energy (u) of the working substance is unchanged by the cyclic process, since internal energy is a property of state and the initial and final state are the same. Then from the first law of thermodynamics: we find: The net heat absorbed (q) is equal to the work done (w) during the cyclic process (or cycle). Reversible transformation - each state of the system is in equilibrium so that a reversal in the direction of an infinitesimal change returns the working substance and the environment to their original states. Heat engine - a device that does work through the agency of heat. € q − w = u2 − u1 € q − w = 0 q = w Consider a cycle of a heat engine in which Q1 heat is absorbed and Q2 heat is rejected. The net heat absorbed is: q = Q1 - Q2 and the work (w) done by the engine is: The efficiency of the engine (h) is defined as: Consider an ideal heat engine as illustrated below: Y - cylinder B - conducting base P - frictionless piston S - nonconducting stand H - infinite warm reservoir C - infinite cold reservoir T1 > T2 Heat is supplied to the working substance when the cylinder (Y) is placed on the warm reservoir (H). Heat is extracted from the working substance when the cylinder (Y) is placed on the cold reservoir (C). Work is done by the working substance when it expands and the piston (P) is pushed outward. € w = q =Q1 −Q2 € η = Work done by the engine Heat absorbed by the working substance = w Q1 = Q1 −Q2 Q1 Carnot’s theorems For a given range of temperatures no engine can be more efficient than a reversible engine. All reversible engines, working between the same range of temperatures, have the same efficiency. For the Carnot cycle the ratio of heat absorbed (Q1) to heat rejected (Q2) is equal to the ratio of the temperature of the warm reservoir (T1) to the temperature of the cold reservoir (T2): What are some examples of real heat engines? What is an atmospheric example of a heat engine? If the Carnot cycle is run in reverse a quantity of heat (Q2) is taken from the cold reservoir and a quantity of heat (Q1) is transferred to the warm reservoir (Q1>Q2). In order for heat to be transferred from the cold to the warm reservoir mechanical work (=Q1 - Q2) must be done on the working substance. What is an example of a reversed heat engine? Another statement of the second law of thermodynamics is “heat cannot of itself (i.e. without the performance of work by some external agency) pass from a cooler to a warmer body in a cyclic process” € Q1 Q2 = T1 T2 Entropy Is heat added to or extracted from the working substance for an isothermal transition from adiabat q1 to q2? The heat associated with this transition is Qrev. If we consider another isothermal transition from adiabat q1 to q2 at a different temperature the amount of heat associated with the transition will be different but the ratio will be the same. We can then use as a measure of the difference between the two adiabats, which is referred to as the difference in entropy. The change in entropy of a system is defined by: and the change in entropy for a unit mass of a substance is given by: Entropy is a function of state of the system and is independent of the path by which the system is brought to that state. For the transition of a system from state 1 to state 2: € Qrev T € Qrev T € dS ≡ dQrev T € ds ≡ dqrev T € s2 − s1 = dqrev T1 2 ∫ Taking the first law of thermodynamics: and the definition of entropy gives: which is a form of the first law of thermodynamics that contains only functions of state. We can relate entropy and potential temperature by taking the first law of thermodynamics expressed as: and rewriting using the equation of state to give: Combining this with Poisson’s equation expressed as: gives Integrating this equation gives: € dq = du + pdα € Tds = du + pdα € dq = cpdT −αdp € dq = cpdT − RT p dp dq T = cp dT T − R dp p € cp dθ θ = cp dT T − R dp p € dq T = cp dθ θ ds = cp dθ θ € s = cp lnθ + constant Gibbs function (G): For a phase change G1 = G2 Differentiation of G gives: From the first law of thermodynamics and the definition of entropy: Using this expression in the equation for dG gives: Since G1 = G2 and dG1 = dG2 then Noting that: gives which is the Clausius-Clapeyron equation. € G = u + esα −Ts € dG = du + esdα +αdes −Tds − sdT € dq = du + pdα = du + esdα =Tds € dG =αdes − sdT € dG1 =α1des − s1dT =α2des − s2dT = dG2 € α1des −α2des = s1dT − s2dT des α1 −α2( ) = dT s1 − s2( ) des dT = s1 − s2( ) α1 −α2( ) = s2 − s1( ) α2 −α1( ) € s2 − s1( ) = ds = dq T = Lv T € des dT = Lv T α2 −α1( ) In the atmosphere a2 >> a1 (the specific volume of water vapor >> the specific volume of liquid water) and using the equation of state for water vapor allows us to write the Clausius-Clapeyron equation as: We can then use this equation to determine changes in es as T varies, by integrating this equation, while assuming that Lv is constant: where es0 is the known saturation vapor pressure at temperature T0. es0 can be determined experimentally, and for T0 = 0 deg C es0 = 611 Pa Also at T0 = 0 deg C = 273.15 K Lv = 2.50x106 J kg-1 Using these values of Lv, T0, and es0 gives: , where: € des dT ≈ Lv Tα2 = Lves RvT 2 € des eses0 (T0 ) es (T ) ∫ = Lv RvT 2 dT T0 T ∫ ln es es0 # $ % & ' ( = − Lv Rv 1 T − 1 T0 # $ % & ' ( = Lv Rv 1 T0 − 1 T # $ % & ' ( es (T) = es0 exp L Rv 1 T0 − 1 T # $ % & ' ( * + , - . / € es T( ) = Aexp −B T # $ % & ' ( € A = es0 exp Lv RvT0 " # $ % & ' = 2.56 ×1011 Pa B = Lv Rv = 5.42 ×103 K Example: Calculate the saturation vapor pressure at the ATOC weather station at SEEC. As the air temperature increases the amount of water vapor required for the air to become saturated increases at an exponential rate. Similarly, as air is cooled the amount of water vapor required for the air to become saturated decreases at an exponential rate. The derivation of the Clausius-Clapeyron equation given above assumed that Lv was constant, but in fact Lv varies with T. From our derivation of the Clausius-Clapeyron equation we know: and a2 >> a1, so Using the equation of state for water vapor (esa2 = RvT) gives: Taking the derivative of Lv with respect to T gives: € Lv = u2 − u1 + es α2 −α1( ) € Lv ≈ u2 − u1 + esα2 € Lv = u2 − u1 + RvT € dLv dT = du2 dT − du1 dT + Rv The dew point temperature (Td) can be used with the Clausius-Clapeyron equation to calculate the vapor pressure (e) with: (units: hPa) The constants e0, b, T1, and T2 are the same as those used when calculating es from T. Example: Calculate the vapor pressure, saturation vapor pressure, and relative humidity from the observed T, Td, and p at the ATOC weather station. Using the Clausius-Clapeyron equation and the relationship between the humidity variables discussed earlier we can convert between any of the following variables: Actual Moisture Content Saturated Moisture Content Td e w q T es ws qs € e = e0 exp b Td −T1( ) Td −T2 # $ % & ' ( Saturation vapor pressure over an ice surface (esi) The saturation vapor pressure over an ice surface can be calculated from the Clausius-Clapeyron equation by replacing Lv, the latent heat of vaporization, with Ls, the latent heat of sublimation. Latent heat of sublimation (Ls): The heat required to convert a unit mass of ice to vapor at constant temperature and pressure (= 2.85x106 J kg-1 K-1) The saturation vapor pressure over an ice surface is given by: , where Combining the expressions for es and esi gives: where Lf is the latent heat of fusion (=3.34x105 J kg-1) € esi = esi0 exp Ls Rv 1 T0 − 1 T # $ % & ' ( ) * + , - . esi = Ai exp −Bi T ) * + , - . € Ai = esi0 exp Ls RvT0 " # $ % & ' = 4.13×1012 Pa B = Ls Rv = 6.18 ×103 K € es T( ) esi T( ) = exp L f RvT0 T0 T −1 # $ % & ' ( ) * + , - . Generalized Statement of the Second Law of Thermodynamics The first part of the second law of thermodynamics states: for a reversible transformation there is no change in the entropy of the universe (where universe refers to a system and its surroundings). Therefore if a system receives heat reversibly the increase in its entropy is exactly equal to the decrease in entropy of the surroundings. In reality all natural transformations are irreversible to some extent. For an irreversible transformation: and there is no simple relationship between the change in entropy of the system and the change in entropy of its surroundings. The second part of the second law of thermodynamics states: the entropy of the universe increases as a result of irreversible transformations The second law of thermodynamics can then be summarized by: DSuniverse = DSsystem + DSsurroundings DSuniverse = 0 for reversible (equilibrium) transformations DSuniverse > 0 for irreversible (spontaneous) transformations In general entropy is a measure of the degree of disorder (randomness) of a system. Therefore irreversible transformations increase the randomness of the universe. € ds ≠ dqirrev T